let m, n be Nat; ( 2 * (m + 1) <= n implies card ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))) = card (Choose (n,m,1,0)) )
defpred S1[ set , set ] means for p being XFinSequence of
for k being Nat st $1 = p & (2 * k) + 1 = min* { N where N is Element of NAT : 2 * (Sum (p | N)) > N } holds
ex r1, r2 being XFinSequence of st
( $2 = r1 ^ r2 & len r1 = (2 * k) + 1 & len r1 = len (p | ((2 * k) + 1)) & p = (p | ((2 * k) + 1)) ^ r2 & ( for o being Nat st o < (2 * k) + 1 holds
r1 . o = 1 - (p . o) ) );
assume A1:
2 * (m + 1) <= n
; card ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))) = card (Choose (n,m,1,0))
A2:
card n = n
by CARD_1:def 2;
A3:
m <= m + m
by XREAL_1:31;
m <= m + 1
by NAT_1:13;
then
2 * m <= 2 * (m + 1)
by XREAL_1:64;
then
2 * m <= n
by A1, XXREAL_0:2;
then
m <= n
by A3, XXREAL_0:2;
then
(card n) choose m > 0
by A2, Th30;
then reconsider W = Choose (n,m,1,0) as non empty finite set by CARD_1:27, CARD_FIN:16;
set Z = Domin_0 (n,(m + 1));
set CH = Choose (n,(m + 1),1,0);
A4:
for x being set st x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))) holds
ex y being set st
( y in W & S1[x,y] )
proof
let x be
set ;
( x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))) implies ex y being set st
( y in W & S1[x,y] ) )
assume A5:
x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))
;
ex y being set st
( y in W & S1[x,y] )
x in Choose (
n,
(m + 1),1,
0)
by A5, XBOOLE_0:def 5;
then consider p being
XFinSequence of
such that A6:
p = x
and A7:
dom p = n
and A8:
rng p c= {0,1}
and A9:
Sum p = m + 1
by CARD_FIN:40;
not
p in Domin_0 (
n,
(m + 1))
by A5, A6, XBOOLE_0:def 5;
then
not
p is
dominated_by_0
by A7, A9, Def2;
then consider o being
Nat such that A10:
(
(2 * o) + 1
= min* { N where N is Element of NAT : 2 * (Sum (p | N)) > N } &
(2 * o) + 1
<= dom p &
o = Sum (p | (2 * o)) &
p . (2 * o) = 1 )
by A8, Th19;
set q =
p | ((2 * o) + 1);
consider r2 being
XFinSequence of
such that A11:
p = (p | ((2 * o) + 1)) ^ r2
by Th5;
rng (p | ((2 * o) + 1)) c= rng p
by RELAT_1:70;
then
rng (p | ((2 * o) + 1)) c= {0,1}
by A8, XBOOLE_1:1;
then consider r1 being
XFinSequence of
such that A12:
len (p | ((2 * o) + 1)) = len r1
and A13:
rng r1 c= {0,1}
and A14:
for
i being
Nat st
i <= len (p | ((2 * o) + 1)) holds
(Sum ((p | ((2 * o) + 1)) | i)) + (Sum (r1 | i)) = 1
* i
and A15:
for
i being
Nat st
i in len (p | ((2 * o) + 1)) holds
r1 . i = 1
- ((p | ((2 * o) + 1)) . i)
by Th29;
take R =
r1 ^ r2;
( R in W & S1[x,R] )
len p = (len r1) + (len r2)
by A12, A11, AFINSQ_1:17;
then A16:
dom R = n
by A7, AFINSQ_1:def 3;
rng r2 c= rng p
by A11, AFINSQ_1:25;
then
rng r2 c= {0,1}
by A8, XBOOLE_1:1;
then
(rng r1) \/ (rng r2) c= {0,1}
by A13, XBOOLE_1:8;
then A17:
rng R c= {0,1}
by AFINSQ_1:26;
(
(p | ((2 * o) + 1)) | (dom (p | ((2 * o) + 1))) = p | ((2 * o) + 1) &
r1 | (dom r1) = r1 )
by RELAT_1:69;
then A18:
(Sum (p | ((2 * o) + 1))) + (Sum r1) = 1
* (len (p | ((2 * o) + 1)))
by A12, A14;
A19:
2
* o < (2 * o) + 1
by NAT_1:13;
then
2
* o c= (2 * o) + 1
by NAT_1:39;
then A20:
(p | ((2 * o) + 1)) | (2 * o) = p | (2 * o)
by RELAT_1:74;
A21:
(2 * o) + 1
c= dom p
by A10, NAT_1:39;
then A22:
dom (p | ((2 * o) + 1)) = (2 * o) + 1
by RELAT_1:62;
A23:
len (p | ((2 * o) + 1)) = (2 * o) + 1
by A21, RELAT_1:62;
then A24:
p | ((2 * o) + 1) = ((p | ((2 * o) + 1)) | (2 * o)) ^ <%((p | ((2 * o) + 1)) . (2 * o))%>
by AFINSQ_1:56;
2
* o in (2 * o) + 1
by A19, NAT_1:44;
then
(p | ((2 * o) + 1)) . (2 * o) = p . (2 * o)
by A23, FUNCT_1:47;
then
Sum (p | ((2 * o) + 1)) = (Sum (p | (2 * o))) + (Sum <%(p . (2 * o))%>)
by A24, A20, AFINSQ_2:55;
then A25:
Sum (p | ((2 * o) + 1)) = o + 1
by A10, AFINSQ_2:53;
m + 1
= (Sum (p | ((2 * o) + 1))) + (Sum r2)
by A9, A11, AFINSQ_2:55;
then
(Sum r1) + (Sum r2) = ((m + 1) - ((2 * o) + 1)) + (2 * o)
by A18, A22, A25;
then
Sum (r1 ^ r2) = m
by AFINSQ_2:55;
hence
R in W
by A16, A17, CARD_FIN:40;
S1[x,R]
thus
S1[
x,
R]
verum
end;
consider F being Function of ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))),W such that
A28:
for x being set st x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))) holds
S1[x,F . x]
from FUNCT_2:sch 1(A4);
W c= rng F
proof
let x be
set ;
TARSKI:def 3 ( not x in W or x in rng F )
assume
x in W
;
x in rng F
then consider p being
XFinSequence of
such that A29:
p = x
and A30:
dom p = n
and A31:
rng p c= {0,1}
and A32:
Sum p = m
by CARD_FIN:40;
set M =
{ N where N is Element of NAT : 2 * (Sum (p | N)) < N } ;
m < m + 1
by NAT_1:13;
then
2
* m < 2
* (m + 1)
by XREAL_1:68;
then
2
* m < n
by A1, XXREAL_0:2;
then
( 2
* (Sum (p | n)) < n &
n in NAT )
by A30, A32, ORDINAL1:def 12, RELAT_1:69;
then A33:
n in { N where N is Element of NAT : 2 * (Sum (p | N)) < N }
;
{ N where N is Element of NAT : 2 * (Sum (p | N)) < N } c= NAT
then reconsider M =
{ N where N is Element of NAT : 2 * (Sum (p | N)) < N } as non
empty Subset of
NAT by A33;
ex
k being
Nat st
(
(2 * k) + 1
= min* M &
Sum (p | ((2 * k) + 1)) = k &
(2 * k) + 1
<= dom p )
proof
set mm =
min* M;
min* M in M
by NAT_1:def 1;
then A34:
ex
o being
Element of
NAT st
(
min* M = o & 2
* (Sum (p | o)) < o )
;
then reconsider m1 =
(min* M) - 1 as
Element of
NAT by NAT_1:20;
A35:
2
* (Sum (p | (min* M))) < m1 + 1
by A34;
A36:
m1 < m1 + 1
by NAT_1:13;
then
m1 c= min* M
by NAT_1:39;
then A37:
(p | (min* M)) | m1 = p | m1
by RELAT_1:74;
min* M <= dom p
by A30, A33, NAT_1:def 1;
then A38:
min* M c= dom p
by NAT_1:39;
then
dom (p | (min* M)) = min* M
by RELAT_1:62;
then
m1 in dom (p | (min* M))
by A36, NAT_1:44;
then A39:
(p | (min* M)) . m1 = p . m1
by FUNCT_1:47;
m1 < m1 + 1
by NAT_1:13;
then
not
m1 in M
by NAT_1:def 1;
then
2
* (Sum (p | m1)) >= m1
;
then A40:
(
Sum <%(p . m1)%> = p . m1 &
(2 * (Sum (p | m1))) + (2 * (p . m1)) >= m1 + 0 )
by AFINSQ_2:53, XREAL_1:7;
reconsider S =
Sum (p | (min* M)) as
Element of
NAT by ORDINAL1:def 12;
take
S
;
( (2 * S) + 1 = min* M & Sum (p | ((2 * S) + 1)) = S & (2 * S) + 1 <= dom p )
A41:
min* M <= dom p
by A30, A33, NAT_1:def 1;
len (p | (min* M)) = m1 + 1
by A38, RELAT_1:62;
then
p | (min* M) = ((p | (min* M)) | m1) ^ <%((p | (min* M)) . m1)%>
by AFINSQ_1:56;
then
Sum (p | (min* M)) = (Sum (p | m1)) + (Sum <%(p . m1)%>)
by A39, A37, AFINSQ_2:55;
hence
(
(2 * S) + 1
= min* M &
Sum (p | ((2 * S) + 1)) = S &
(2 * S) + 1
<= dom p )
by A41, A40, A35, NAT_1:9;
verum
end;
then consider k being
Nat such that A42:
(2 * k) + 1
= min* M
and A43:
Sum (p | ((2 * k) + 1)) = k
and A44:
(2 * k) + 1
<= dom p
;
set k1 =
(2 * k) + 1;
consider q being
XFinSequence of
such that A45:
p = (p | ((2 * k) + 1)) ^ q
by Th5;
rng (p | ((2 * k) + 1)) c= rng p
by RELAT_1:70;
then
rng (p | ((2 * k) + 1)) c= {0,1}
by A31, XBOOLE_1:1;
then consider r being
XFinSequence of
such that A46:
len r = len (p | ((2 * k) + 1))
and A47:
rng r c= {0,1}
and A48:
for
i being
Nat st
i <= len (p | ((2 * k) + 1)) holds
(Sum ((p | ((2 * k) + 1)) | i)) + (Sum (r | i)) = 1
* i
and A49:
for
i being
Nat st
i in len (p | ((2 * k) + 1)) holds
r . i = 1
- ((p | ((2 * k) + 1)) . i)
by Th29;
set rq =
r ^ q;
A50:
dom (r ^ q) = (len (p | ((2 * k) + 1))) + (len q)
by A46, AFINSQ_1:def 3;
A51:
m = k + (Sum q)
by A32, A43, A45, AFINSQ_2:55;
dom (r ^ q) = (len (p | ((2 * k) + 1))) + (len q)
by A46, AFINSQ_1:def 3;
then A52:
dom (r ^ q) = dom p
by A45, AFINSQ_1:def 3;
(
(p | ((2 * k) + 1)) | (dom (p | ((2 * k) + 1))) = p | ((2 * k) + 1) &
r | (dom r) = r )
by RELAT_1:69;
then A53:
(Sum (p | ((2 * k) + 1))) + (Sum r) = 1
* (len (p | ((2 * k) + 1)))
by A46, A48;
rng q c= rng p
by A45, AFINSQ_1:25;
then
rng q c= {0,1}
by A31, XBOOLE_1:1;
then
(rng r) \/ (rng q) c= {0,1}
by A47, XBOOLE_1:8;
then A54:
rng (r ^ q) c= {0,1}
by AFINSQ_1:26;
A55:
(2 * k) + 1
c= dom p
by A44, NAT_1:39;
then A56:
len (p | ((2 * k) + 1)) = (2 * k) + 1
by RELAT_1:62;
then A57:
(r ^ q) | ((2 * k) + 1) = r
by A46, AFINSQ_1:57;
A58:
((2 * k) + 1) + 1
> (2 * k) + 1
by NAT_1:13;
then A59:
(2 * k) + 1
< 2
* (Sum r)
by A43, A53, A56;
A60:
2
* (Sum r) > (2 * k) + 1
by A43, A53, A56, A58;
then consider j being
Nat such that A61:
(
(2 * j) + 1
= min* { N where N is Element of NAT : 2 * (Sum ((r ^ q) | N)) > N } &
(2 * j) + 1
<= dom (r ^ q) &
j = Sum ((r ^ q) | (2 * j)) &
(r ^ q) . (2 * j) = 1 )
by A54, A57, Th6, Th19;
set j1 =
(2 * j) + 1;
A62:
len ((p | ((2 * k) + 1)) ^ q) = (len (p | ((2 * k) + 1))) + (len q)
by AFINSQ_1:def 3;
not
r ^ q is
dominated_by_0
by A60, A57, Th6;
then A63:
not
r ^ q in Domin_0 (
n,
(m + 1))
by Th24;
set rqj =
(r ^ q) | ((2 * j) + 1);
Sum (r ^ q) = (Sum r) + (Sum q)
by AFINSQ_2:55;
then
r ^ q in Choose (
n,
(m + 1),1,
0)
by A30, A43, A53, A56, A52, A54, A51, CARD_FIN:40;
then A64:
r ^ q in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))
by A63, XBOOLE_0:def 5;
then consider r1,
r2 being
XFinSequence of
such that A65:
F . (r ^ q) = r1 ^ r2
and A66:
len r1 = (2 * j) + 1
and A67:
(
len r1 = len ((r ^ q) | ((2 * j) + 1)) &
r ^ q = ((r ^ q) | ((2 * j) + 1)) ^ r2 )
and A68:
for
i being
Nat st
i < (2 * j) + 1 holds
r1 . i = 1
- ((r ^ q) . i)
by A28, A61;
A69:
dom (r ^ q) = (len r1) + (len r2)
by A67, AFINSQ_1:def 3;
then A70:
len (r1 ^ r2) = len ((p | ((2 * k) + 1)) ^ q)
by A50, A62, AFINSQ_1:17;
reconsider K =
{ N where N is Element of NAT : 2 * (Sum ((r ^ q) | N)) > N } as non
empty Subset of
NAT by A61, NAT_1:def 1;
(r ^ q) | ((2 * k) + 1) = r
by A46, A56, AFINSQ_1:57;
then
(2 * k) + 1
in K
by A59;
then A71:
(2 * k) + 1
>= (2 * j) + 1
by A61, NAT_1:def 1;
then
(2 * j) + 1
c= (2 * k) + 1
by NAT_1:39;
then A72:
(p | ((2 * k) + 1)) | ((2 * j) + 1) = p | ((2 * j) + 1)
by RELAT_1:74;
(2 * j) + 1
in K
by A61, NAT_1:def 1;
then A73:
ex
N being
Element of
NAT st
(
N = (2 * j) + 1 & 2
* (Sum ((r ^ q) | N)) > N )
;
(Sum ((p | ((2 * k) + 1)) | ((2 * j) + 1))) + (Sum (r | ((2 * j) + 1))) = ((2 * j) + 1) * 1
by A48, A56, A71;
then
2
* (Sum (r | ((2 * j) + 1))) = (2 * ((2 * j) + 1)) - (2 * (Sum (p | ((2 * j) + 1))))
by A72;
then
((2 * j) + 1) + (((2 * j) + 1) - (2 * (Sum (p | ((2 * j) + 1))))) > (2 * (Sum (p | ((2 * j) + 1)))) + (((2 * j) + 1) - (2 * (Sum (p | ((2 * j) + 1)))))
by A46, A56, A71, A73, AFINSQ_1:58;
then
(2 * j) + 1
> 2
* (Sum (p | ((2 * j) + 1)))
by XREAL_1:6;
then
(2 * j) + 1
in M
;
then
(2 * j) + 1
>= (2 * k) + 1
by A42, NAT_1:def 1;
then A74:
(2 * j) + 1
= (2 * k) + 1
by A71, XXREAL_0:1;
A75:
len ((p | ((2 * k) + 1)) ^ q) = len (r ^ q)
by A50, AFINSQ_1:def 3;
now let i be
Nat;
( i < len (r1 ^ r2) implies (r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i )assume A76:
i < len (r1 ^ r2)
;
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . inow per cases
( i < len r1 or i >= len r1 )
;
suppose A77:
i < len r1
;
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . ithen A78:
(
i in dom r1 &
r1 . i = 1
- ((r ^ q) . i) )
by A66, A68, NAT_1:44;
A79:
i in len r1
by A77, NAT_1:44;
A80:
(
len r1 = len (p | ((2 * k) + 1)) &
i in NAT )
by A55, A66, A74, ORDINAL1:def 12, RELAT_1:62;
then A81:
r . i = 1
- ((p | ((2 * k) + 1)) . i)
by A49, A79;
(
((p | ((2 * k) + 1)) ^ q) . i = (p | ((2 * k) + 1)) . i &
(r ^ q) . i = r . i )
by A46, A79, A80, AFINSQ_1:def 3;
hence
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i
by A81, A78, AFINSQ_1:def 3;
verum end; suppose A82:
i >= len r1
;
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . ithen A83:
((p | ((2 * k) + 1)) ^ q) . i = q . (i - (len r))
by A46, A56, A66, A74, A70, A76, AFINSQ_1:19;
(
(r1 ^ r2) . i = r2 . (i - (len r1)) &
(r ^ q) . i = q . (i - (len r)) )
by A46, A56, A66, A74, A70, A75, A76, A82, AFINSQ_1:19;
hence
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i
by A67, A70, A75, A76, A82, A83, AFINSQ_1:19;
verum end; end; end; hence
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i
;
verum end;
then A84:
r1 ^ r2 = (p | ((2 * k) + 1)) ^ q
by A69, A50, A62, AFINSQ_1:9, AFINSQ_1:17;
r ^ q in dom F
by A64, FUNCT_2:def 1;
hence
x in rng F
by A29, A45, A65, A84, FUNCT_1:3;
verum
end;
then A85:
rng F = W
by XBOOLE_0:def 10;
A86:
F is one-to-one
proof
let x be
set ;
FUNCT_1:def 4 for b1 being set holds
( not x in proj1 F or not b1 in proj1 F or not F . x = F . b1 or x = b1 )let y be
set ;
( not x in proj1 F or not y in proj1 F or not F . x = F . y or x = y )
assume that A87:
x in dom F
and A88:
y in dom F
and A89:
F . x = F . y
;
x = y
x in Choose (
n,
(m + 1),1,
0)
by A87, XBOOLE_0:def 5;
then consider p being
XFinSequence of
such that A90:
p = x
and A91:
dom p = n
and A92:
rng p c= {0,1}
and A93:
Sum p = m + 1
by CARD_FIN:40;
not
p in Domin_0 (
n,
(m + 1))
by A87, A90, XBOOLE_0:def 5;
then
not
p is
dominated_by_0
by A91, A93, Def2;
then consider k1 being
Nat such that A94:
(
(2 * k1) + 1
= min* { N where N is Element of NAT : 2 * (Sum (p | N)) > N } &
(2 * k1) + 1
<= dom p &
k1 = Sum (p | (2 * k1)) &
p . (2 * k1) = 1 )
by A92, Th19;
y in Choose (
n,
(m + 1),1,
0)
by A88, XBOOLE_0:def 5;
then consider q being
XFinSequence of
such that A95:
q = y
and A96:
dom q = n
and A97:
rng q c= {0,1}
and A98:
Sum q = m + 1
by CARD_FIN:40;
not
q in Domin_0 (
n,
(m + 1))
by A88, A95, XBOOLE_0:def 5;
then
not
q is
dominated_by_0
by A96, A98, Def2;
then consider k2 being
Nat such that A99:
(
(2 * k2) + 1
= min* { N where N is Element of NAT : 2 * (Sum (q | N)) > N } &
(2 * k2) + 1
<= dom q &
k2 = Sum (q | (2 * k2)) &
q . (2 * k2) = 1 )
by A97, Th19;
A100:
len q = n
by A96;
reconsider M =
{ N where N is Element of NAT : 2 * (Sum (q | N)) > N } as non
empty Subset of
NAT by A99, NAT_1:def 1;
set qk =
q | ((2 * k2) + 1);
consider s1,
s2 being
XFinSequence of
such that A101:
F . y = s1 ^ s2
and A102:
len s1 = (2 * k2) + 1
and A103:
len s1 = len (q | ((2 * k2) + 1))
and A104:
q = (q | ((2 * k2) + 1)) ^ s2
and A105:
for
i being
Nat st
i < (2 * k2) + 1 holds
s1 . i = 1
- (q . i)
by A28, A88, A95, A99;
A106:
len q = (len (q | ((2 * k2) + 1))) + (len s2)
by A104, AFINSQ_1:17;
then A107:
len q = len (s1 ^ s2)
by A103, AFINSQ_1:17;
reconsider K =
{ N where N is Element of NAT : 2 * (Sum (p | N)) > N } as non
empty Subset of
NAT by A94, NAT_1:def 1;
set pk =
p | ((2 * k1) + 1);
consider r1,
r2 being
XFinSequence of
such that A108:
F . x = r1 ^ r2
and A109:
len r1 = (2 * k1) + 1
and A110:
len r1 = len (p | ((2 * k1) + 1))
and A111:
p = (p | ((2 * k1) + 1)) ^ r2
and A112:
for
i being
Nat st
i < (2 * k1) + 1 holds
r1 . i = 1
- (p . i)
by A28, A87, A90, A94;
assume
x <> y
;
contradiction
then consider i being
Nat such that A113:
i < len p
and A114:
p . i <> q . i
by A90, A91, A95, A96, A106, AFINSQ_1:9;
A115:
len p = (len (p | ((2 * k1) + 1))) + (len r2)
by A111, AFINSQ_1:17;
then A116:
len p = len (r1 ^ r2)
by A110, AFINSQ_1:17;
now per cases
( k1 = k2 or k1 > k2 or k1 < k2 )
by XXREAL_0:1;
suppose A117:
k1 = k2
;
contradictionnow per cases
( i < len (p | ((2 * k1) + 1)) or i >= len (p | ((2 * k1) + 1)) )
;
suppose A118:
i < len (p | ((2 * k1) + 1))
;
contradictionthen
i in len (p | ((2 * k1) + 1))
by NAT_1:44;
then A119:
(
r1 . i = (r1 ^ r2) . i &
s1 . i = (s1 ^ s2) . i )
by A109, A110, A102, A117, AFINSQ_1:def 3;
(
r1 . i = 1
- (p . i) &
s1 . i = 1
- (q . i) )
by A109, A110, A112, A105, A117, A118;
hence
contradiction
by A89, A108, A101, A114, A119;
verum end; suppose A120:
i >= len (p | ((2 * k1) + 1))
;
contradictionthen A121:
(s1 ^ s2) . i = s2 . (i - (len (p | ((2 * k1) + 1))))
by A91, A109, A110, A96, A102, A107, A113, A117, AFINSQ_1:19;
(
p . i = r2 . (i - (len (p | ((2 * k1) + 1)))) &
q . i = s2 . (i - (len (p | ((2 * k1) + 1)))) )
by A91, A109, A110, A111, A102, A103, A104, A100, A113, A117, A120, AFINSQ_1:19;
hence
contradiction
by A89, A108, A110, A101, A116, A113, A114, A120, A121, AFINSQ_1:19;
verum end; end; end; hence
contradiction
;
verum end; suppose A122:
k1 > k2
;
contradiction
len s1 <= len p
by A91, A96, A103, A106, NAT_1:11;
then A123:
len s1 c= dom p
by NAT_1:39;
2
* k2 < 2
* k1
by A122, XREAL_1:68;
then A124:
len s1 < len r1
by A109, A102, XREAL_1:6;
then
(s1 ^ s2) | (len s1) = r1 | (len s1)
by A89, A108, A101, AFINSQ_1:58;
then A125:
s1 = r1 | (len s1)
by AFINSQ_1:57;
A126:
for
k being
Nat st
k < len (q | ((2 * k2) + 1)) holds
(q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k
proof
let k be
Nat;
( k < len (q | ((2 * k2) + 1)) implies (q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k )
assume A127:
k < len (q | ((2 * k2) + 1))
;
(q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k
A128:
k in len s1
by A103, A127, NAT_1:44;
then A129:
k in (dom q) /\ (len s1)
by A91, A96, A123, XBOOLE_0:def 4;
k in (dom p) /\ (len s1)
by A123, A128, XBOOLE_0:def 4;
then A130:
p . k = (p | (len (q | ((2 * k2) + 1)))) . k
by A103, FUNCT_1:48;
A131:
k < len r1
by A103, A124, A127, XXREAL_0:2;
then A132:
r1 . k = 1
- (p . k)
by A109, A112;
k in dom r1
by A131, NAT_1:44;
then
k in (dom r1) /\ (len s1)
by A128, XBOOLE_0:def 4;
then A133:
r1 . k = (r1 | (len s1)) . k
by FUNCT_1:48;
s1 . k = 1
- (q . k)
by A102, A103, A105, A127;
hence
(q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k
by A102, A125, A132, A133, A129, A130, FUNCT_1:48;
verum
end;
(2 * k2) + 1
in M
by A99, NAT_1:def 1;
then A134:
ex
N being
Element of
NAT st
(
(2 * k2) + 1
= N & 2
* (Sum (q | N)) > N )
;
len (q | ((2 * k2) + 1)) = len (p | (len (q | ((2 * k2) + 1))))
by A103, A123, RELAT_1:62;
then
q | ((2 * k2) + 1) = p | (len (q | ((2 * k2) + 1)))
by A126, AFINSQ_1:9;
then
len (q | ((2 * k2) + 1)) in K
by A102, A103, A134;
hence
contradiction
by A94, A109, A103, A124, NAT_1:def 1;
verum end; suppose A135:
k1 < k2
;
contradiction
len r1 <= len q
by A91, A110, A96, A115, NAT_1:11;
then A136:
len r1 c= dom q
by NAT_1:39;
2
* k1 < 2
* k2
by A135, XREAL_1:68;
then A137:
len r1 < len s1
by A109, A102, XREAL_1:6;
then
(r1 ^ r2) | (len r1) = s1 | (len r1)
by A89, A108, A101, AFINSQ_1:58;
then A138:
r1 = s1 | (len r1)
by AFINSQ_1:57;
A139:
for
k being
Nat st
k < len (p | ((2 * k1) + 1)) holds
(p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k
proof
let k be
Nat;
( k < len (p | ((2 * k1) + 1)) implies (p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k )
assume A140:
k < len (p | ((2 * k1) + 1))
;
(p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k
A141:
k in len r1
by A110, A140, NAT_1:44;
then A142:
k in (dom p) /\ (len r1)
by A91, A96, A136, XBOOLE_0:def 4;
k in (dom q) /\ (len r1)
by A136, A141, XBOOLE_0:def 4;
then A143:
q . k = (q | (len (p | ((2 * k1) + 1)))) . k
by A110, FUNCT_1:48;
A144:
k < len s1
by A110, A137, A140, XXREAL_0:2;
then A145:
s1 . k = 1
- (q . k)
by A102, A105;
k in dom s1
by A144, NAT_1:44;
then
k in (dom s1) /\ (len r1)
by A141, XBOOLE_0:def 4;
then A146:
s1 . k = (s1 | (len r1)) . k
by FUNCT_1:48;
r1 . k = 1
- (p . k)
by A109, A110, A112, A140;
hence
(p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k
by A109, A138, A145, A146, A142, A143, FUNCT_1:48;
verum
end;
(2 * k1) + 1
in K
by A94, NAT_1:def 1;
then A147:
ex
N being
Element of
NAT st
(
(2 * k1) + 1
= N & 2
* (Sum (p | N)) > N )
;
len (p | ((2 * k1) + 1)) = len (q | (len (p | ((2 * k1) + 1))))
by A110, A136, RELAT_1:62;
then
p | ((2 * k1) + 1) = q | (len (p | ((2 * k1) + 1)))
by A139, AFINSQ_1:9;
then
len (p | ((2 * k1) + 1)) in M
by A109, A110, A147;
hence
contradiction
by A110, A99, A102, A137, NAT_1:def 1;
verum end; end; end;
hence
contradiction
;
verum
end;
dom F = (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))
by FUNCT_2:def 1;
then
(Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))),W are_equipotent
by A86, A85, WELLORD2:def 4;
hence
card ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))) = card (Choose (n,m,1,0))
by CARD_1:5; verum